Proof of Implicit Function Theorem: special case

Question

I was reading this pdf online on the Implicit Function Theorem (pg 18-19). I did not understand the last line : Why "continuity of $f$ and uniqueness of $y$ implies continuity of $g$"?

Answer 1

You have: $0=f(x+h,g(x+h))-f(x,g(x))$ by the multivariable Lagrange theorem for all small $h$ we have $\theta\in (0,1)$ such that:

$$f(x+h,g(x+h))-f(x,g(x))=f_x(x + \theta h, g(x) + θ(g(x + h) − g(x)))h+ f_y(x + \theta h, g(x) +\theta (g(x + h) − g(x)))(g(x + h) − g(x))$$ then by re-arranging we have: $$|g(x+h)-g(x)|=\frac{|f_x(x + \theta h, g(x) + θ(g(x + h) − g(x)))|}{|f_y(x + \theta h, g(x) +\theta (g(x + h) − g(x)))|} |h|$$

this gives you continuity since the RHS is bounded by $M|h|$. Now that you know $g$ is continuous. Use the above equation with absolute value to get that the derivative at $x$ is $\frac{-f_x(x,g(x))}{f_y(x,g(x))}$.

If you need more details I guess I can add them.